Talk at Tsinghua University 2012,3,16 Nonlinear Schrödinger Equation BAOXIANG WANG School of Mathematical Sciences, Peking University 1 1 33
1. Schrödinger E. Schrödinger (1887-1961) E. Schrödinger, (1887, 8 1961,1), born in Vienna, Nobel Prize Winner in 1933, he made a fundamental contribution in quantum mechanics, so called Schrödnger equation. Schrödinger is one of the greatest physicians in the twenty century. He has a wide interests. 5WHAT IS LIFE?6is a famous book of Schrödinger in molecular biology, where you can find his great ambitions. Schrödinger is also a poet and he wrote many poems. 1 2 33
2. Linear Schrödinger equation iu t + ( + V )u = 0, u(0, x) = u 0 (x), (1) u(t, x) : R R n C u 0 : R n C i = 1 u t = u t u = 2 u 2 x 1 +... + 2 u 2 x n V (x), known function plays a fundamental role in quantum mechanics just as the Newton laws in the classical mechanics. 1 3 33
From physical point of views, de Broglie conjectured that Any microparticle has the wave-particle duality. Schrödinger calculated that de Broglie wave [ ] i ψ = A exp (p r Et) satisfies the following equation iψ t + 2 2µ ψ = (E p2 /2µ)ψ. (2) Along this way, Schrödinger realized that a general wave function u should satisfies (1). Now we have the following two fundamental assumptions in quantum mechanics: (I) Wave function u completely describes the moving states of particles (II) The evolution of a wave function satisfies (1). 1 4 33
Nonlinear Schrödinger equation (NLS) In many branches, such as plasma physics, optics, deep water wave, even in economic theory, nonlinear Schrödinger equation was discovered: iu t + u = λ u α u, u(0, x) = u 0 (x), (3) If λ > 0, (3) is said to be defocusing NLS If λ < 0, (3) is said to be focusing NLS 1 5 33
NLS is one of the most important equation in physics. To understand NLS from mathematical point of view, it seems that i plays a crucial role, let us recall a proverb of Leibniz: Imaginary number is the magical flight of the spirit of the god, it is almost between the existence and nonexistence of amphibians. Jê þ2 Ûœ1, A 0u3ÚØ3üa 1 6 33
2.1. Conservation Laws NLS has two basic conservation laws: Mass and Energy. Taking the inner product of (3) with u, and then considering the imaginary part, we have Proposition 1. Let u be a suitable smooth solution of (3), we have 1 2 u(t) 2 L 2 = 1 2 u 0 2 L 2 (Mass) (4) Taking the inner product of (3) with u t and then preserving the real part, we obtain Denote 1 d 2 dt u(t) 2 L + 2 E(u(t)) = 1 2 u(t) 2 L 2 + λ d 2 + α dt u(t) 2+α L = 0 2+α λ 2 + α u(t) 2+α L 2+α. Proposition 2. Let u be a suitable smooth solution of (3). We have E(u(t)) = E(u 0 ). (Energy) (5) 1 7 33
In E(u(t)), 1 2 u(t) 2 L = kinetic energy 2 λ 2+α u(t) 2+α L =potential energy. 2+α In the defocusing case (λ > 0), both kinetic and potential energies are positive, it is a good case. In the focusing case (λ < 0), kinetic energy is positive, and potential energy is negative, so it is a bad case. Roughly speaking, good case means that the global solutions exists, and bad case means that the solution will blow up at finite time. NLS also satisfies some other conservation laws, let us mention the pseudoconformal invariant law: Proposition 3. Let u be the smooth solution of (3). We have 1 2 (x + 2it )u(t) 2 L + 8λt2 2 2 + α u(t) 2+α L 2+α + 4λ t (4 nα) 2 + α 0 Ginibre and Velo, J. Funct. Anal. 32 (1979) 1-72. s u(s) 2+α 2+α ds = xu 0 2 L 2. (pseudo-conformal invariant law) 1 8 33
2.2. Critical power Let us understand the critical power from the scaling. If u solves NLS (3), so does u ν (t, x) = ν 2/α u(ν 2 t, νx) (6) solve iv t + v = λ v α v, v(0, x) = ν 2/α u 0 (νx), (7) Let us write f Ḣ1 = f L 2. One can calculate the norm of ν 2/α u 0 (νx) in Ḣ1 : u ν (0, ) Ḣ1 = ν 2/α u 0 (ν ) L 2 = ν 2/α+1 ( u 0 )(νx) L 2 = ν 2/α+1 n/2 u 0 L 2 1 9 33
We consider the following question: One needs the condition = ν 2/α+1 n/2 u 0 Ḣ1 Can we take some α so that u ν (0, ) Ḣ1 is invariant with respect to ν > 0? ν 2/α+1 n/2 = 1 α = 4 n 2. Definition 4. For NLS (3), α = 4 n 2 is said to be the critical power in Ḣ1. 0 < α < 4 n 2 (0 < α < for n = 1, 2) is said to be the subcritical power in Ḣ1. 1 10 33
Obviously, The above procedure can be extended to the space L 2, Ḣ k, f Ḣk = k f L 2: Definition 5. For NLS (3), α = 4 n 2k is said to be the critical power in Ḣk. 0 < α < 4 n 2k (α < for n 2k) is said to be the subcritical power in Ḣk. In particular, α = 4 n said to be the critical power in L2. 0 < α < 4 n said to be the subcritical power in L 2. Ḣ s : The above definition can be generalized to the fractional Sobolev spaces f Ḣs = F 1 ξ s F f L 2 Definition 6. For NLS (3), α = 4 n 2s is said to be the critical power in Ḣs. α < 4 n 2s (α < for n 2s) is said to be the subcritical power in Ḣs. 1 11 33
3. Global Well Posedness For NLS, which is said to be locally well-posed in Ḣs if for any initial value u 0 Ḣ s, there exists a unique solution in C(0, T ; Ḣs ) for some T > 0, moreover, the solution map u 0 u preserves the regularity from Ḣr to C(0, T ; Ḣr ) ( r 0) and it is at least a continuous mapping from Ḣs to C(0, T ; Ḣs ). Furthermore, if T =, then we say that it is globally well-posed. 1 12 33
Global well posedness in L 2 Theorem 7. (Tsutsumi Y. 1984) Let u 0 L 2, α < 4 n i.e., it is a subcritical power in L 2 = (3) is globally well posed in L 2. Theorem 8. (Cazenave T., Weissler F.B., 1990) Let u 0 L 2 be sufficiently small, α = 4 n, i.e., it is a critical power in L2 = (3) is globally well posed in L 2. In the defocusing L 2 -critical case, Theorem 9. (Killip, Tao, Visan 2009, Dodson 2010,2011) Let λ > 0. Let u 0 L 2, α = 4 n, i.e., it is a critical power in L2 = (3) is globally well posed in L 2. In the focusing L 2 -critical case, the question is related to the ground state solution for an elliptic equation: Q + Q 1+4/n = Q One can find that for the above Q, u = e it Q is a solution of the focusing NLS (λ = 1). Merle F. showed that u 0 2 > Q 2, there are some solutions of the focusing NLS blows up at finite time. So, it is natural to conjecture that 1 13 33
Problem 10. Let λ = 1, u 0 L 2 with u 0 2 < Q 2, α = 4 n, i.e., it is a critical power in L 2 = (3) is globally well posed in L 2? We only know partial answer to this question. Visan M. (2008) showed it is true in 2D radial case. 1 14 33
Global well posedness in H 1 Recall that u Ḣ1 r = u L r, H 1 r = L r Ḣ1 r, H 1 = L 2 Ḣ1. Denote 2 =, n = 1, 2 (8) 2 = 2n n 2, n 3 (9) Theorem 11. (Kato T. 1987) Let u 0 H 1 α < 2 2 = defocusing (λ > 0) NLS (3) is globally well posed in H 1. Theorem 12. (Bourgain J. 1999, Tao T., Visan, 2005, 2008) Let u 0 Ḣ1 α = 4 n 2 = defocusing (λ > 0) NLS (3) is globally well posed in Ḣ1. When α > 4/(n 2), we only know the existence od weak solutions: Theorem 13. (School of Lions J.L., 1950 1960) Let u 0 H 1 L 2+α α > 4 n 2 = defocusing (λ > 0) NLS (3) has at least one weak solution u C w (R, H 1 (R n ) L 2+α (R n )). 1 15 33
Problem 14. For α > 4 n 2 posed? and λ > 0, is the weak solution unique? global well For the above question, it seems far from to have an answer. We do not know how to solve it using the known method. 1 16 33
For the focusing case λ < 0, it is related to the elliptic equation where we have a radial solution The blow up result is W + W 4/(n 2) W = 0, W (x) = ) 1 n/2 (1 + x 2 n(n 2) Let n 3 and u 0 Ḣ1 with E(u 0 ) < E(W ) and u 0 2 W 2. Assume also that either xu 0 L 2 or u 0 H 1 is radial. Then the corresponding solution u to focusing NLS blows up in finite time. So, it is natural to conjecture that Let n 3 and u 0 Ḣ1, E(u 0 ) > E(W ) (+...); or E(u 0 ) < E(W ) and u 0 2 < W 2. Then the focusing NLS is globally well posed. 1 17 33
Now only partial answer is known Theorem 15. (Kenig, Merle, 2007, Killip Visan 2008) Let n 3 and u 0 Ḣ1 E(u 0 ) < E(W ) and u 0 2 < W 2. We further assume that u 0 is radial for n = 3, 4. Then the focusing NLS is globally well posed. 1 18 33
Well posedness in H s (= L 2 Ḣs ) Let us recall that for s > 0, u Ḣs r = ( ) s/2 u L r, Hr s = L r Ḣs r, H s = L 2 Ḣs. Theorem 16. (Cazenave T., Weissler F.B., 1990) Let 0 < s < n/2, u 0 H s, u 0 Ḣs sufficiently small, α is critical in Ḣs, i.e., α = 4/(n 2s) = NLS (3) is globally well posed in H s. Problem 17. For the defocusing NLS (3), can we remove the condition that u 0 Ḣ s is sufficiently small? There is no full answer even in 1D. 1 19 33
4. Definition 18. (Definition of the scattering from H s H s ). Let u 0 Hs, and u (t) be the solution of iu t + u = 0, u (0, x) = u 0. Assume that NLS iu t + u = u α u has a unique solution u satisfying u(t) u (t) H s 0, t. Moreover, we assume that u + (t) satisfies iu + t + u + = 0, u + (0, x) = u + 0, and u(t) u + (t) H s 0, t +. Then we say that S : u 0 u+ 0 from Hs to H s is a scattering operator. We have u (t) = S(t)u 0 = e it u 0. One can construct the scattering operator in the following way: t u(t) S(t)u 0 = i S(t τ)[ u α u](τ)dτ. u(t) S(t)u + 0 = i S(t τ)[ u α u](τ)dτ. t 1 20 33
Theorem 19. (Nakanishi 1999; Ginibre, Velo 1989) Let λ > 0, 4/n < α < 2 2 (see (8)) = the sacttering operator of NLS S : H 1 H 1 is a homeomorphism. Theorem 20. (Tao, Visan 2005, 2008 Ann. Math.) Let λ > 0, n 3, α = 2 2(i.e., H 1 critical case)= the sacttering operator of NLS S : H 1 H 1 is a homeomorphism. Theorem 21. (Glassey, 1973) 0 < α < 2/n, = the sacttering operator of NLS does NOT exist. i.e., u(t) S(t)u 0 H 1 0, t for some u 0. Problem 22. In the case 2/n < α 4/n, there is no complete result. For any status u 0, the existence of the sacttering operator is OPEN! 1 21 33
Theorem 23. α > 4/(n 2), α = 4/(n 2s) = S : B δ (0, H s ) := {u H s : u Ḣs δ} H s. for some small δ > 0. (This result is independent of the energy) Problem 24. For any status u 0 Hs, it is still OPEN! Nonlinear term for NLS with exponential growth iu t + u = (e u m 1)u, u(0, x) = u 0 (x). (10) Theorem 25. (Wang and Hao; 2005) n = 1, m 6 = the sacttering operator of NLS (10) S : homeomorphism. H 1 H 1 is a 1 22 33 Problem 26. In 2D case, there is a similar result. If n 3, there is no any result for NLS (10).
5. Blow up for NLS We consider the Blow up solutions for NLS in the focusing case: Recall that the energy is iu t + u + u α u = 0, u(0, x) = u 0 (x), (11) E(u(t)) = 1 2 u(t) 2 L 2 1 2 + α u(t) 2+α L 2+α. We have known that 0 < α < 4 n = (11) is global well posed in L2. Theorem 27. α 4/n, E(u 0 ) < 0 = the solution of (11) blows up in finite time: There exists T > 0 such that 1 23 33 lim sup t T xu(t) L 2 =.
Blow up rate, dynamical behavior (whether the Blow up rate is stable with a small perturbation to the initial data) are not very clear for a long time, Blow up solutions of (11) seem RATHER complicated. Theorem 28. Merle F. Ann. of Math. 2005,2006 n = 1, α = 4, E(u 0 ) < c, c 1 < u 0 2 L 2 < c 1 + ε = the solution of (11) blows up at finite time T > 0 and x u L 2 C ln ln(t t) 1/4 T t 1/2. Problem 29. For n 2, α = 4/n, we do not know if the above Theorem still holds. When α > 4/n there is no any corresponding results. 1 24 33
Concentration of blow up solutions. Theorem 30. (Bourgain J. 1999, Merle 1998) u 0 H 1, α = 4/n, Let the solution of (11) blows up at T > 0, then we have for some a R n, u(t) L 2 ( x a T t 1/2 ) c Problem 31. We do not know if the concentration is localized in a bounded region in R n. So, Bourgain J. ask the following questions: If or not the concentration point a move away to infinity? We only know partial answer to this question. 1 25 33
Theorem 32. (Wang 2006) n = 1, α = 4, E(u 0 ) < c, c 1 < u 0 2 L 2 < c 1 + ε = (11) blows up at finite time T > 0, and moreover, there exists at most finite many a i R satisfying lim sup t T u(t) L 2 ( x a i T t 1/2 ) c Problem 33. n 2, α = 4/n, there is no complete result. 1 26 33
6. (DNLS) iu t + u + λ 1 ( u 2 u) + ( λ 2 u) u 2 = 0, u(0, x) = u 0 (x), (12) where i, are the same ones as before. λi is complex vector. It seems that DNLS is more difficult than NLS, one must find a way to handle the derivative in nonlinearity. 1 27 33
One can generalize DNLS into the following iu t + u = F (u, ū, u, ū), u(0, x) = u 0 (x), (13) u : R R n C; n u = x 2 i, (14) i=1 = ( x1,..., xn ), F (z) = P (z 1,..., z 2n+2 ) = c β z β, c β C, (15) m, M N, c β C. m+1 β <M +1 1 28 33
Theorem 34. (Local Solution) (Chihara, 1999, Math. Ann. ) Let n 2, s > 4 + n/2. u 0 H s. 2 m M <. Then (13) is local well posed. Theorem 35. ( with small data) (Ozawa, Zhai 2008, Ann I H Poincare) Let n 3, s > n/2 + 2. u 0 H s. 2 m M. Re F/ ( u) = (θ( u 2 )), θ C 2, θ(0) = 0 (H) Then (13) is globally well posed. 1 29 33
7. Non-elliptic Derivative Schrödinger Equation aries from the many-body system. iu t + ± u = F (u, ū, u, ū), u(0, x) = u 0 (x), (16) u : R R n C; n ± u = ε i x 2 i, ε i {1, 1}, (17) i=1 = ( x1,..., xn ), F (z) = P (z 1,..., z 2n+2 ) = c β z β, c β C, (18) m, M N, c β C. m+1 β <M +1 1 30 33
Theorem 36. (Local Well Posedness) (Kenig-Ponce-Vega, 1998, 2004, Invent. Math.) Let n 2, s n/2. u 0 H s. 2 m M <. Then (13) is locall well posed in C([0, T ]; H s ) for some T > 0. Theorem 37. (Global solution for small data) (Wang, Han, Huang, Ann I H Poincare 2009) Let n 2, 2 m <, m > 4/n. u 0 H (n+3)/2 be sufficiently small. Then (16) is globally well posed. Moreover, the solution is scattering. Problem 38. Is derivative nonlinear Schrödinger equation (12) global well posed for any data? It is open in higher spatial dimensions n 2. 1 31 33
8. M t = M ± M, M(0, x) = M 0 (x) Making a stereographic projection ( ) 2Imu 2Reu M = u 2 1 1 + u 2, 1 + u 2, 1 + u 2 u satisfies a non-elliptic derivative NLS i iu t + ± u = λ ε i( xi u) 2 u 1 + u 2 = λ ( 1) k k=0 n ε i ( xi u) 2 u 2k u. (19) Problem 39. is a difficult problem, the global well posedness has been open for many years. If the initial data are sufficiently small in Ḣn/2, the global solution in the elliptic case was recently obtained by Benjanaru, Ionescu, Kenig, Tartaru (Ann Math, 2011), in the nonelliptic case, the global existence is covered by Theorem 37.. i=1 1 32 33
Thank you! 1 33 33